Problem: $h(n) = 3n^{2}-2n$ $g(n) = 2n^{3}+5n^{2}+5n-5+5(f(n))$ $f(n) = 4n^{2}+3(h(n))$ $ f(g(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(0)$ . Then we'll know what to plug into the outer function. $g(0) = 2(0^{3})+5(0^{2})+(5)(0)-5+5(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = 4(0^{2})+3(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = 3(0^{2})+(-2)(0)$ $h(0) = 0$ That means $f(0) = 4(0^{2})+(3)(0)$ $f(0) = 0$ That means $g(0) = 2(0^{3})+5(0^{2})+(5)(0)-5+(5)(0)$ $g(0) = -5$ Now we know that $g(0) = -5$ . Let's solve for $f(g(0))$ , which is $f(-5)$ $f(-5) = 4(-5)^{2}+3(h(-5))$ To solve for the value of $f$ , we need to solve for the value of $h(-5)$ $h(-5) = 3(-5)^{2}+(-2)(-5)$ $h(-5) = 85$ That means $f(-5) = 4(-5)^{2}+(3)(85)$ $f(-5) = 355$